if $E\ge_{\text{Nak}}0$, strictly in one point, $H^{n,q}(X,E)=0$, for $q\ge 1$;

if $E\le_{\text{Nak}}0$, strictly in one point, $H^{p,0}(X,E)=0$, for $p<n$.

On the other hand, Nakano's positivity is not well-behaved with respect to taking duals: we have that $E$ is Griffiths positive if and only if $E^*$ is Griffiths negative, but if we take $H\to\mathbb P^n$ to be the vector bundle of rank $n$ defined by
$$
0\to\mathcal O(-1)\to\underline{\mathbb C}^{n+1}\to H\to 0,
$$
where $\underline{\mathbb C}^{n+1}$ si the trivial vector bundle over $\mathbb P^n$ with fiber $\mathbb C^{n+1}$, then $H$ is Griffiths (semi)positive and $H^*$ is Nakano (semi)negative but $H$ is neither Nakano (semi)positive nor Nakano (semi)negative.

If we look to short exact sequences of vector bundles
$$
0\to S\to E\to Q\to 0,
$$
then the Nakano's negativity of $E$ implies the Nakano's negativity of $S$, but nothing can be said about the Nakano's positivity of $Q$ when $E$ is Nakano positive (the desired property holds if we look instead to Griffiths' positivity).

Of course Nakano's positivity implies the Griffiths' one; the following "partial converse" is due to Demailly and Skoda:

Nakano (semi-)positivity is not an algebraic notion. It implies Griffiths positivity, which implies ampleness. It is known that Griffiths positivity does not imply Nakano positivity (an example is the tangent bundle of the complex projective space), but it is not known whether ampleness implies Griffiths positivity.

For more on this see 6.1.D of Lazarsfeld's book. The fact that the tangent bundle of the complex projective space is not Nakano positive follows from a vanishing theorem p.97 of ibid. that fails for it.